To model the scenario where Lorraine hits a golf ball, we can use a quadratic equation of the form \( f(x) = ax^2 + bx + c \). Given that the ball lands 60 feet away and reaches a maximum height of 30 feet, we can denote the horizontal distance from Lorraine to the point where the ball lands as the x-coordinate and the height of the ball as the y-coordinate.
- Finding the equation:
Since the ball reaches its maximum height (30 feet) at its vertex, we can assume that the vertex form of the parabola is \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. From the problem statement:
- The vertex \( (h, k) \) is \( (30, 30) \) (the maximum height occurs in the horizontal range).
- The ball hits the ground (height = 0) at \( x = 0 \) feet and \( x = 60 \) feet.
We know it can be expressed in vertex form and that it will pass through the point (0, 0) when the ball is at the starting point.
Now, we rewrite the equation in standard vertex form: \[ f(x) = a(x - 30)^2 + 30 \] To find the value of \( a \), we can use one of the intercepts, say \( (0, 0) \): \[ 0 = a(0 - 30)^2 + 30 \] \[ 0 = 900a + 30 \] \[ 900a = -30 \] \[ a = -\frac{1}{30} \]
Now, we can express the full equation: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \] Expanding this gives: \[ f(x) = -\frac{1}{30}(x^2 - 60x + 900) + 30 \] \[ f(x) = -\frac{1}{30}x^2 + 2x - 30 + 30 \] \[ f(x) = -\frac{1}{30}x^2 + 2x \]
So, the equation is: \[ f(x) = -\frac{1}{30}x^2 + 2x \]
- Finding the height of the ball when it is 15 feet away from Lorraine:
Now we can find \( f(15) \): \[ f(15) = -\frac{1}{30}(15^2) + 2(15) \] \[ f(15) = -\frac{1}{30}(225) + 30 \] \[ f(15) = -7.5 + 30 \] \[ f(15) = 22.5 \]
Therefore:
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The equation is: \[ f(x) = -\frac{1}{30}x^2 + 2x \]
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The height of the ball when it is 15 feet away from Lorraine is: \( 22.5 \) feet in the air.
Final answers:
- \[ f(x) = -\frac{1}{30}x^2 + 2x \]
- Height = \( 22.5 \) feet.
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